(1).
f(x) = |x − 5| + 3
f(2) = |2 − 5| + 3 = |−3| + 3 = 3 + 3 = 6
(2).
f(x) = 8 − 3|x + 2|
f(−6) = 8 − 3|−6 + 2| = 8 − 3|−4| = 8 − 3·4 = 8 − 12 = − 4
(3).
For f(x) = |x − h| + k the turning point is (h, k), so:
f(x) = |x − 2| + 3 ⇒ the turning point is: (2, 3)